The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is $60 \mathrm{~cm}$, the length of the closed pipe will be :
JEE MAIN 2024, Diffcult
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$\frac{\lambda}{4}=\mathrm{L}_1 \quad 2\left(\frac{\lambda}{2}\right)=\lambda$
$\mathrm{v}=\mathrm{f} \lambda \quad \mathrm{f}_2=\frac{2 \mathrm{v}}{2 \mathrm{~L}_2}$
$\mathrm{v}=\mathrm{f}_1\left(4 \mathrm{~L}_1\right) \quad \mathrm{f}_2=\frac{\mathrm{v}}{\mathrm{L}_2}$
${v}=\mathrm{f}_1\left(4 \mathrm{~L}_1\right)$
${f}_2=\frac{\mathrm{v}}{\mathrm{L}_2}$
${f}_1=\frac{\mathrm{v}}{4 \mathrm{~L}_1}$
${f}_1=\mathrm{f}_2$
$\frac{\mathrm{V}}{4 \mathrm{~L}_1}=\frac{\mathrm{V}}{\mathrm{L}_2}$
$\Rightarrow \mathrm{L}_2=4 \mathrm{~L}_1$
$60=4 \times \mathrm{L}_1$
${~L}_1=15 \mathrm{~cm} $
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